Find the sum of each of the following series: (i) tan⁻¹(1/(x² + x + 1)) + tan⁻¹(1/(x² + 3x + 3)) + tan⁻¹(1/(x² + 5x + 7)) + tan⁻¹(1/(x² + 7x + 13)) +... up to n terms. (ii) tan⁻¹(1/3) + tan⁻¹(2/9) +... + tan⁻¹(2^(n-1)/(1 + 2^(2n-1))) +... up to infinite terms. (iii) sin⁻¹(1/sqrt(2)) + sin⁻

Question

Find the sum of each of the following series: (i) tan⁻¹(1/(x² + x + 1)) + tan⁻¹(1/(x² + 3x + 3)) + tan⁻¹(1/(x² + 5x + 7)) + tan⁻¹(1/(x² + 7x + 13)) +... up to n terms. (ii) tan⁻¹(1/3) + tan⁻¹(2/9) +... + tan⁻¹(2^(n-1)/(1 + 2^(2n-1))) +... up to infinite terms. (iii) sin⁻¹(1/sqrt(2)) + sin⁻¹((sqrt(2) - 1)/sqrt(6)) +... + sin⁻¹((sqrt(n) - sqrt(n-1))/sqrt(n(n+1))) +... up to infinite terms.

Accepted Answer

(i) tan⁻¹(x+n) - tan⁻¹(x); (ii) pi/4; (iii) pi/2. (i) The k-th term tan⁻¹(1/(x² + (2k-1)x + (k² - k + 1))) can be written as tan⁻¹(x + k) - tan⁻¹(x + k - 1). Summing k = 1 to n telescopes to tan⁻¹(x + n) - tan⁻¹(x). (ii) The k-th term tan⁻¹(2^(k-1)/(1 + 2^(2k-1))) = tan⁻¹(2^k) - tan⁻¹(2^(k-1)). Summing to infinity gives lim tan⁻¹(2ⁿ) - tan⁻¹(2⁰) = pi/2 - pi/4 = pi/4. (iii) The k-th term sin⁻¹((sqrt(k) - sqrt(k-1))/sqrt(k(k+1))) = sin⁻¹(sqrt(k/(k+1))) - sin⁻¹(sqrt((k-1)/k)), telescoping to lim sin⁻¹(sqrt(n/(n+1))) - sin⁻¹(0) = pi/2 - 0 = pi/2.

Solution

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